Answer :
Certainly! Let's work through this step by step.
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First Part: Cross between Heterozygous Male (Ww) and Homozygous Recessive Female (ww)
When we cross a heterozygous male (Ww) with a homozygous recessive female (ww), we need to construct a Punnett square to determine the genotypes of the offspring.
1. Construct the Punnett Square:
[tex]\[ \begin{array}{|c|c|c|} \hline & W & w \\ \hline w & Ww & ww \\ \hline w & Ww & ww \\ \hline \end{array} \][/tex]
2. Determine the Probabilities:
- From the Punnett square:
- Ww (heterozygous) appears in two squares.
- ww (homozygous recessive) appears in two squares.
- Thus, the probability of having heterozygous (Ww) offspring is calculated as follows:
[tex]\[ \text{Probability of heterozygous offspring} = \frac{\text{Number of Ww outcomes}}{\text{Total outcomes}} = \frac{2}{4} = \frac{1}{2} = 0.5 \][/tex]
So, if a heterozygous male with the genotype [tex]\( Ww \)[/tex] is mated with a homozygous recessive female of genotype [tex]\( ww \)[/tex], there is a chance that 0.5 of the offspring will be heterozygous.
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Second Part: Cross between Heterozygous (Ww) and Homozygous Dominant (WW)
When crossing a heterozygous individual (Ww) with a homozygous dominant (WW), we again construct the Punnett square.
1. Construct the Punnett Square:
[tex]\[ \begin{array}{|c|c|c|} \hline & W & w \\ \hline W & WW & Ww \\ \hline W & WW & Ww \\ \hline \end{array} \][/tex]
2. Determine the Probabilities:
- From the Punnett square:
- WW (homozygous dominant) appears in two squares.
- Ww (heterozygous) appears in two squares.
- ww (homozygous recessive) does not appear in any square (zero squares).
- Thus, the probability of having a homozygous recessive (ww) offspring is:
[tex]\[ \text{Probability of homozygous recessive offspring} = \frac{\text{Number of ww outcomes}}{\text{Total outcomes}} = \frac{0}{4} = 0 \][/tex]
So, if the heterozygous [tex]\( Ww \)[/tex] is crossed with a homozygous dominant [tex]\( WW \)[/tex], then the probability of having a homozygous recessive offspring is 0.
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To summarize the correct answer:
- If a heterozygous male with the genotype [tex]\( Ww \)[/tex] is mated with a homozygous recessive female of genotype [tex]\( ww \)[/tex], there is a chance that 0.5 of the offspring will be heterozygous.
- If the heterozygous [tex]\( Ww \)[/tex] is crossed with a homozygous dominant [tex]\( WW \)[/tex], then the probability of having a homozygous recessive offspring is 0.
---
First Part: Cross between Heterozygous Male (Ww) and Homozygous Recessive Female (ww)
When we cross a heterozygous male (Ww) with a homozygous recessive female (ww), we need to construct a Punnett square to determine the genotypes of the offspring.
1. Construct the Punnett Square:
[tex]\[ \begin{array}{|c|c|c|} \hline & W & w \\ \hline w & Ww & ww \\ \hline w & Ww & ww \\ \hline \end{array} \][/tex]
2. Determine the Probabilities:
- From the Punnett square:
- Ww (heterozygous) appears in two squares.
- ww (homozygous recessive) appears in two squares.
- Thus, the probability of having heterozygous (Ww) offspring is calculated as follows:
[tex]\[ \text{Probability of heterozygous offspring} = \frac{\text{Number of Ww outcomes}}{\text{Total outcomes}} = \frac{2}{4} = \frac{1}{2} = 0.5 \][/tex]
So, if a heterozygous male with the genotype [tex]\( Ww \)[/tex] is mated with a homozygous recessive female of genotype [tex]\( ww \)[/tex], there is a chance that 0.5 of the offspring will be heterozygous.
---
Second Part: Cross between Heterozygous (Ww) and Homozygous Dominant (WW)
When crossing a heterozygous individual (Ww) with a homozygous dominant (WW), we again construct the Punnett square.
1. Construct the Punnett Square:
[tex]\[ \begin{array}{|c|c|c|} \hline & W & w \\ \hline W & WW & Ww \\ \hline W & WW & Ww \\ \hline \end{array} \][/tex]
2. Determine the Probabilities:
- From the Punnett square:
- WW (homozygous dominant) appears in two squares.
- Ww (heterozygous) appears in two squares.
- ww (homozygous recessive) does not appear in any square (zero squares).
- Thus, the probability of having a homozygous recessive (ww) offspring is:
[tex]\[ \text{Probability of homozygous recessive offspring} = \frac{\text{Number of ww outcomes}}{\text{Total outcomes}} = \frac{0}{4} = 0 \][/tex]
So, if the heterozygous [tex]\( Ww \)[/tex] is crossed with a homozygous dominant [tex]\( WW \)[/tex], then the probability of having a homozygous recessive offspring is 0.
---
To summarize the correct answer:
- If a heterozygous male with the genotype [tex]\( Ww \)[/tex] is mated with a homozygous recessive female of genotype [tex]\( ww \)[/tex], there is a chance that 0.5 of the offspring will be heterozygous.
- If the heterozygous [tex]\( Ww \)[/tex] is crossed with a homozygous dominant [tex]\( WW \)[/tex], then the probability of having a homozygous recessive offspring is 0.
